Solve 5x^2-17*5x-10=0

$$5 x ^ { 2 } - 17 \cdot 5 x - 10 = 0$$

Solve for x

$$x = \frac{3 \sqrt{33} + 17}{2} \approx 17.11684397$$

$$x=\frac{17-3\sqrt{33}}{2}\approx -0.11684397$$

Steps Using the Quadratic Formula

Multiply 17 and 5 to get 85.

$$5x^{2}-85x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -85 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-85\right)±\sqrt{\left(-85\right)^{2}-4\times 5\left(-10\right)}}{2\times 5}$$

Square -85.

$$x=\frac{-\left(-85\right)±\sqrt{7225-4\times 5\left(-10\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-85\right)±\sqrt{7225-20\left(-10\right)}}{2\times 5}$$

Multiply -20 times -10.

$$x=\frac{-\left(-85\right)±\sqrt{7225+200}}{2\times 5}$$

Add 7225 to 200.

$$x=\frac{-\left(-85\right)±\sqrt{7425}}{2\times 5}$$

Take the square root of 7425.

$$x=\frac{-\left(-85\right)±15\sqrt{33}}{2\times 5}$$

The opposite of -85 is 85.

$$x=\frac{85±15\sqrt{33}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{85±15\sqrt{33}}{10}$$

Now solve the equation x=\frac{85±15\sqrt{33}}{10} when ± is plus. Add 85 to 15\sqrt{33}.

$$x=\frac{15\sqrt{33}+85}{10}$$

Divide 85+15\sqrt{33} by 10.

$$x=\frac{3\sqrt{33}+17}{2}$$

Now solve the equation x=\frac{85±15\sqrt{33}}{10} when ± is minus. Subtract 15\sqrt{33} from 85.

$$x=\frac{85-15\sqrt{33}}{10}$$

Divide 85-15\sqrt{33} by 10.

$$x=\frac{17-3\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{33}+17}{2} x=\frac{17-3\sqrt{33}}{2}$$

Steps for Completing the Square

Multiply 17 and 5 to get 85.

$$5x^{2}-85x-10=0$$

Add 10 to both sides. Anything plus zero gives itself.

$$5x^{2}-85x=10$$

Divide both sides by 5.

$$\frac{5x^{2}-85x}{5}=\frac{10}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{85}{5}\right)x=\frac{10}{5}$$

Divide -85 by 5.

$$x^{2}-17x=\frac{10}{5}$$

Divide 10 by 5.

$$x^{2}-17x=2$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=2+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=2+\frac{289}{4}$$

Add 2 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{297}{4}$$

Factor x^{2}-17x+\frac{289}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{297}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{297}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{3\sqrt{33}}{2} x-\frac{17}{2}=-\frac{3\sqrt{33}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{3\sqrt{33}+17}{2} x=\frac{17-3\sqrt{33}}{2}$$